Monotonicity for excited random walk in high dimensions

Transcription

1 Monotonicity for excite ranom walk in high imensions Remco van er Hofsta Mark Holmes March, 2009 Abstract We prove that the rift θ, β) for excite ranom walk in imension is monotone in the excitement parameter β [0, ], when is sufficiently large. We give an explicit criterion for monotonicity involving ranom walk Green s functions, an use rigorous numerical upper bouns provie by Hara [8] to verify the criterion for 9. Introuction In this paper we stuy excite ranom walk, where the ranom walker has a rift in the irection of the first component each time the walker visits a new site. It was shown that this process has ballistic behaviour when 2 in [4, 3, 4]. A nontrivial strong law of large numbers SLLN) can then be obtaine for 2 using renewal techniques see for example [5, 6]). For =, it is known that ERW is recurrent an iffusive [7] except in the trivial case β =. Aitional results on one-imensional multi)-excite ranom walks can be foun in [, 2, 3, 6]. In [0], a perturbative expansion was introuce an use to prove a weak law of large numbers an a central limit theorem for excite ranom walk in imensions > 5 an > 8 respectively, with sufficiently small excitement parameter. More recently, [5] explicitly prove a SLLN an establishe a functional central limit theorem in imensions 2. Inclue in [0] is an explicit representation of the rift in terms of the expansion coefficients. In this paper we use this representation, together with explicit simple ranom walk Green s function bouns [8, 9] to prove that in imensions 9, the rift for excite ranom walk is strictly) increasing in the excitement parameter β.. Main results The main result of this paper is the following theorem. Department of Mathematics an Computer Science, Einhoven University of Technology, P.O. Box 53, 5600 MB Einhoven, The Netherlans. rhofsta@win.tue.nl Department of Statistics, The University of Aucklan, Private Bag 9209, Aucklan 42, New Zealan. mholmes@stat.aucklan.ac.nz

2 Theorem. Monotonicity of the spee). For all 9, an β [0, ], the rift for excite ranom walk in imension with excitement parameter β is strictly increasing in β. We are also able to show that for 8, there exists β 0 ) such that the rift for ERW is strictly increasing in β [0, β 0 ]. Simulations [5] suggest that the limiting variance of the first coorinate is not monotone in the excitement parameter β in 2 imensions. We expect that using the approach introuce in this paper we can show that the variance is monotone ecreasing in β when the imension is taken sufficiently high. By [0], the variance of the first coorinate is equal to σβn 2 + o)) for some asymptotic variance σβ, 2 an base on our methos, we expect that σβ 2 = 2 β 2 + β 2 O 3 ), showing that, in sufficiently high, β σβ 2 is ecreasing. Although we only consier the once-excite ranom walk in this paper, the general multiexcite ranom walk can be hanle with very minor moifications, yieling a result at least as strong as Theorem.. A large part of the methoology in this paper can be applie more generally. See [2] for an application of the methos an results in this paper to the case where the rift on subsequent visits to a site is in the opposite irection to that inuce by the excitement parameter on the first visit. Given the present context, another natural example is a ranom walk in an environment that is ranom in the first few coorinates only, with the expecte rift inuce by the environment enote by β. Some progress has been mae in this irection [] making use of the fact that a SLLN has been prove for general versions of such ranom walks in ranom environment in [6]. We first introuce some notation. A nearest-neighbour ranom walk path η is a sequence {η i } for which η i Z an η i+ η i is a nearest-neighbour of the origin for all i 0. For a general nearest-neighbour path η with η 0 = 0, we write p η i x i, x i+ ) for the conitional probability that the walk steps from η i = x i to x i+, given the history of the path η i = η 0,..., η i ). We write, for β [0, ], p β x) = + βe x I { x =},.) 2 where e =, 0,..., 0), an x y is the inner-prouct between x an y. Thus, p β is the transition probability for a ranom walk having a rift β when stepping in the first coorinate. We write ω n for the n-step path of excite ranom walk ERW), an Q for the law of { ω n } n=0, i.e., for every n-step nearest-neighbour path η n, Q ω n = η n ) = n p η i η i, η i+ ),.2) where p 0 0, η ) = p β η ) is the probability to jump to η in the first step, an p η i η i, η i+ ) = p 0 η i+ η i )I {ηi η i } + p β η i+ η i )[ I {ηi η i }],.3) where I {ηi η i } enotes the inicator that η i = η j for some 0 j i. In wors, the ranom walker gets excite each time he/she visits a new site, an when the ranom walk is excite, it has a positive rift in the irection of the first coorinate. For a escription in terms of cookies, see [6]. 2

3 2 An overview of the proof an the expansion In this section we recall some results an notation from [0]. If η an ω are two paths of length at least j an m respectively an such that η j = ω 0, then the concatenation η j ω m is efine by η j ω m ) i = { ηi when 0 i j, ω i j when j i m + j. 2.) Given η m such that Q ω m = η m ) > 0, we efine a probability measure Q ηm on paths starting from η m by specifying its value on particular cyliner sets in a consistent manner) as follows Q ηm ω n = µ n ) n p ηm µ i µ i, µ i+ ), 2.2) an extening the measure to all finite-imensional cyliner sets in the natural consistent) way. Then 2.2) is also Q ω m+n = η m µ n ω m = η m ). In [0], a perturbative expansion was erive for the two-point function c n x) = Qω n = x), giving rise to a recursion relation of the form c n+ x) = y Z p β y)c n x y) + n+ m=2 y Z π m y)c n+ m x y). 2.3) This expansion was use to prove a law of large numbers an central limit theorem for ERW. We next iscuss the coefficients π m y) an some results of this expansion. The expansion coefficients. Let N, an for i 0, let ω i) j i + be a path of length j i + Z +, where, by convention, j 0 = 0. Then efine N = p ωn ) + ωn) ω p N) ) ω N), ω N) +), 2.4) which epens on ω N ) + an ω N) + although this epenence is suppresse in the notation). The ifference 2.4) is ientically zero when the histories ω N ) + ω N) an ω N) give the same transition probabilities to go from ω N) when ω N) π N) has alreay been visite by ω N ) + but not by ω N) N = βe ω N) + ω N) [ ) 2 to ω N) +. For excite ranom walk, N is non-zero precisely, so that I N) {ω j / ω N ) N j ω N) N } I {ω N) ] / ω N) } β 2 I I β {ω N) + =ωn) j ±e } {ω N) N j ω N ) N j \ ω N) N } 2 I I. {ω N) + =ωn) j ±e } {ω N) N j ω N ) N j } N Define A m,n = {j,..., ) Z N + : N l= j l = m N }, A N = m A m,n an m x, y) = j A m,n ω 0) ω ) j + ω N) + I {ω N) =x,ω N) + =y}p βω 0) N ) n= j n n i n=0 2.5) p ωn ) j n + ωn) in ω n) i n, ω n) i n+). 2.6) 3

4 Then we efine π m x, y) = N= π N) m x, y), π N) x, y) = m π N) m x, y), an π m y) = π N) N= x Z m x, y). 2.7) Note that the quantities π m N) epen on β. Note further that are all zero when N + > m, an that all of the above quantities π m N) x, y) = 0, 2.8) y Z since ω N) N = 0 see also [0, 6.0)]). + The importance of these quantities is given by [0, Proposition 3.], which states that if lim nm=2 n x Z xπ mx) exists an n ω n converges in probability to θ, then θβ, ) = yp β y) + yπ m y) = βe y Z m=2 y Z + m=2 y Z yπ m y). 2.9) Strategy of the proof of Theorem.. We shall explicitly ifferentiate the right han sie of 2.9), an prove that this erivative is positive for all β [0, ], when 9. From 2.9) an using 2.7) 2.8), we have so that yπ m y) = y Z x,y Z y x)π m x, y), 2.0) θβ, ) = βe + m=2 N= x,y Z y x)π N) m x, y). 2.) Letting ϕ N) m x, y) = sums, we then have β πn) m x, y) an assuming that the limit can be taken through the infinite θ β β, ) = e + N= m=2 Since ϕ N) m x, y) 0 unless x y =, we have that x,y Z y x)ϕ N) m x, y). 2.2) θ β β, ) e ϕ N) m x, y). 2.3) N= m=2 x,y Z We conclue that θ θ β, ), which is the first coorinate of β, ), is positive for any β at which β β N= m=2 x,y Z ϕn) m x, y) <. This is what we shall prove in the remainer of this paper, which is organise as follows. In Section 3, we start by proving bouns on π m N). These bouns will be crucially use to prove bouns on ϕ N) m in Section 4. The results in Section 4 are use in Section 5 to prove Theorem.. 4

5 3 Boun on π Before proceeing to the proof of Theorem., we prove a new boun on x,y Z m π N) m x, y). The proof of this new boun makes use of Lemma 3. below. Let P enote the law of simple symmetric ranom walk in imensions, starting at the origin, an let D x) = I { x =} /2) be the simple ranom walk step istribution. We will make use of the convolution of functions, which is efine for absolutely summable functions f, g on Z by f g)x) = y Z fy)gx y). 3.) Let f k enote the k-fol convolution of f with itself, an let G x) = k=0 D k x) enote the Green s function for this ranom walk. We shall sometimes make use of the representation G i x) = k=0 m i :m + +m i =k D m + +m i ) x) = k=0 k + i )! P ω k = x), for i. 3.2) i )!k! Note that G i x) < if an only if > 2i. We shall often abbreviate G i let = G i 0). For i 0, ) i+g i+) E i ) = sup v) δ 0,v ). 3.3) v Z Lemma 3. Diagrammatic bouns for ERW). For excite ranom walk, uniformly in u Z an η m, for i 0, j=0 j= Q ηm ω j = u) i! ) i+g i+), 3.4) Q ηm ω j = u) i!e i ). 3.5) Proof. Define an increasing sequence of ranom variables N j, j 0, by letting j N j be the number of steps that the walk ω j takes in the first coorinate. Observe that inepenently of η, N j has a Binomialj, q ) istribution, where q = )/. If we consier ω j as the initial position an first j steps of an infinite walk ω, then the sequence {N j } j 0 is a ranom walk on Z + taking i.i.. steps that are either + or 0 with probability q an q respectively. The ranom time that such a walk spens at any level l has a Geometric istribution with parameter q. Thus, writing P for the law of {N j } j=0, we obtain that for every i 0, so that, for m l, PN j = l) = j=m l!j l)! ql q ) j l = q i PN j = l) = q i+) l + i)! PN j+i = l + i), l! l + i)!. 3.6) l! 5

6 Given u = u,..., u ) Z, we write u := u 2, u 3,..., u ) Z. To prove 3.4), note that j=0 j Q ηm ω j = u) = Q ηm ω j = u N j = l)pn j = l) j=0 P ω l = u ηm) PN j = l) j=l q i+) sup P ω l = v) v Z l + i)!. 3.7) l! By 3.2), 3.7) is equal to i!q i+) sup v Z G i+) v). By [9, Lemma B.3], the supremum occurs at v = 0. Since q = / ), this proves 3.4). The boun 3.5) is prove similarly. Inee, for i 0, we can write j= Q ηm ω j = u) sup v Z = sup v Z = sup v Z =i! sup v Z P ω l = v) j=l [ P ω l = v) j=l q i+) i+) q PN j = l) PN j = l) δ 0,l i!pn 0 = 0) ) l + i)! P ω l = v) i!δ 0,v l! G i+) v) δ 0,v ), 3.8) since PN 0 = 0) = an P ω l = v)δ 0,l = δ 0,v, an following the steps in 3.7) above. Define a = ) 2 G ) Proposition 3.2 Bouns on the expansion coefficients for ERW). For N =, x,y Z m π ) m x, y) β E 0 ), an, for N 2, x,y Z Given η m an z j+, efine m π N) m x, y) β N ) G E )a N ) z j+ ) = p ηm z j z j, z j+ ) p z j z j, z j+ ) ) I {z0 =η m}. 3.) We will use the following lemma to prove Proposition 3.2. ]) 6

7 Lemma 3.3 Ingreients for bouns on lace expansion coefficients). For any η s, j=0 z j+ z j+ ) j=0j + ) z j+ j= z j+ z j+ ) j=j + ) z j+ j p ηs z i z i, z i+ ) sβ G, 3.2) j z j+ ) p ηs z i z i, z i+ ) sβa, 3.3) j p ηs z i z i, z i+ ) sβ E 0), 3.4) j z j+ ) p ηs z i z i, z i+ ) sβ E ). 3.5) Proof. As in 2.5), the left han sie of 3.2) is boune above by j p ηs z i β z i, z i+ ) I {zj η s } I {zj+ =z j=0 z j 2 j ±e } 3.6) z j+ β j p ηs z i z i, z i+ ) I {zj η s }, j=0 z j since only two terms contribute to the rightmost sum in 3.6). From 2.2), this is equal to β Q ηs ω j = z j )I {zj η s } = β Q ηs ω j η s ) β s Q ηs ω j = η l ). 3.7) j=0 z j j=0 j=0 The inequality 3.2) then follows from 3.4) with i = 0. The inequality 3.3) is obtaine similarly, using 3.4) with i = at the last step, while 3.4) an 3.5) are obtaine using 3.5) with i = 0 an i = respectively at the last step. Proof of Proposition 3.2. It follows from 2.6) that x,y Z m π N) m x, y) is boune by p β ω 0) ) j = ω 0) ω ) j + j ) p ω0) ω) i ω ) i, ω ) i + i = =0 ω N) + N p ωn ) + ωn) i N ω N) i N, ω N) i N +), i N = 3.8) where the sums over j k, k 2 are all from 0 to. Note that can only be non-zero if j is o so in particular, non-zero). We procee by using Lemma 3.3 to successively boun the sums over j k of this expression, beginning with the sum over. If N = then we use 3.4) with s = to boun this sum by sβ E 0), an then p ω 0) β ω 0) ) = gives the result. If N > then we use 3.2) with s = + on the sum over, followe by repeate applications of 3.3) with s = j k + on the sums over j k with k = N,..., 2 respectively, then 3.5) with s = on the sum over j an again the result follows since p ω 0) β ω 0) ) =. Since the spee is known to exist [5], the following corollary is an easy consequence of [0, Propositions 3. an 6.] together with Proposition 3.2, an the fact that a 6 < since G 2 5 < 5 2 /6 [9]. 7

8 Corollary 3.4 Formula for the spee of ERW). For all 6 an β [0, ], θβ, ) = lim E[ω n+ ω n ] = βe n + m=2 x Z xπ m x). 3.9) In fact, the first equality in Corollary 3.4 hols for all 2 since the law µ n of the cookie environment as viewe by the ranom walker at time n is known to converge see e.g. [5]) an E[ω n+ ω n ] = E [ E[ω n+ ω n ω n ] ] [ ] βe = E I {ω n / ω n } = βe [ Pωn ω n ) ], 3.20) where the right han sie converges as n since Pω n ω n ) is the µ n -measure of the event that the cookie at the origin is absent. 4 The ifferentiation step To verify the exchange of limits in 2.2), it is sufficient to prove that x,y Zy x)πn) m x, y) is absolutely summable in m an N note that for every m an N the summations over x an y are finite) an that m=2 N= sup β [0,] x,y Zy x)ϕn) m x, y) <. By Proposition 3.2 an the fact that y x = for x, y nearest neighbours, the first conition hols provie that βa <. 4.) In fact we will see later on that this inequality for β = is sufficient to also establish the secon conition. We now ientify ϕ m N) x, y). Recall 2.6). Then we can write ϕ N) m x, y) = ϕ m N,) x, y) + ϕ N,2) m x, y) + ϕ N,3) m x, y), 4.2) where by Leibniz rule), ϕ m N,) x, y), ϕ m N,2) x, y) an ϕ N,3) m x, y) arise from ifferentiating p β ω 0) ), Nn= jn i n=0 p ωn ) j n + ωn) in ω n) i n, ω n) i n+) an Nn= n, respectively, with respect to β. Observe that if η m = x l then β p ηm β x l, x) = e x x l )I {xl / η m } 2 an hence, using I A I A C = I A C c β Clearly then β I { x xl =} = I {x l / η m } 2 we have I{x xl =e } I {x xl = e } ), 4.3) p η m β x l, x) p ωn ηm β x l, x) ) = 2 I ) {x l / η m,x l ω n } I{x xl =e } I {x xl = e }. 4.4) p η m β x l, x) p ωn ηm β x l, x) ) 2 I ) {x l ω n \ η m } I{x xl =e } + I {x xl = e }. 4.5) Let ρ N) be the quantity obtaine by replacing p β ω 0) ) in 2.6) with 2) I 0) {ω a boun =±e } on its erivative) an by bouning n by n for all n =,..., N. 8

9 For k =,..., N, let γ N) k be the quantity obtaine from 2.6) by bouning n by n for ω k) i k, ω k) i k +) with the following boun on its all n =,..., N an by replacing j k i k =0 p ωk ) j k + ωk) i k erivative j k I {ω k) l+ ωk) l =±e } 2 j k i k = 0 i k l p ωk ) j k + ωk) i k ω k) i k, ω k) i k +). 4.6) Similarly, let χ N) k be obtaine by replacing k in 2.6) by 2) I I a k) {ω j ω k ) k j } {ω k) k j k + ωk) j =±e } k boun on its erivative) an by bouning n for n k by n. Letting γ N) = N k= γ N) k an χ N) = N k= χ N) k, we obtain m x,y Z ϕ N,) m x, y) ρ N), m ϕ N,2) m x, y) γ N), an x,y Z m x,y Z ϕ N,3) We shall boun each of these terms separately, in Lemmas 4., 4.4 an 4.2 below. Lemma 4. Bouns on ρ N) ). For N =, ρ ) 2 βe 0 ), an, for N 2, m x, y) χ N). 4.7) ρ N) β N G E ) 2 ) an ) Proof. This is exactly the same as the proof of Proposition 3.2 except that at the very last step we use 2 I = {ω 0) =±e }. 4.9) ω 0) Lemma 4.2 Bouns on χ N) ). For N =, χ ) E 0 ), an, for N 2, χ N) Nβ N G E ) ) an ) Proof. Proceeing exactly as in the proof of Proposition 3.2, except that the boun on k is missing the β term, we obtain χ ) k E 0 ) an, for N 2, χ N) k β N ) G E )a N 2. 4.) The resulting boun on χ N), which is simply β N times 3.0)) is then easily obtaine by summing over k from to N. Before proceeing to the boun on γ N), we first nee a new lemma similar to Lemma 3.. Lemma 4.3. Let Q l, η s enote the law of a self-interacting ranom walk ω with history η s, where the transition probabilities are those of an ERW with history η s, except that Q l, η s ω l+ = ω l + e ω l ) = Q l, η s ω l+ = ω l e ω l ) = ) 9

10 Then, for all i 0, j= j Q l, η s ω j = u) i + )! ) i+2 G i+2). 4.3) Proof. Since one of the j steps is a simple ranom walk step in the first coorinate, the number of steps in the other coorinates has a Binomialj, ) istribution. Thus, j= j Q l, η s ω j = u) sup j j v Z j= l= k=0 = sup v Z k=0 = sup v Z PN j = k)p ω k = v) P ω k = v) j j=k+ P ω k = v) k=0 r=k PN j = k) r + i + )! PN r = k). 4.4) r! Now procee as in the proof of Lemma 3. to obtain the result. Define 2 ɛ) = ) G G E ) ) 2 G ) Lemma 4.4 Bouns on γ N) ). For N =, 2, γ ) N 3, β ) 2 G 2, γ 2) β 2 ɛ) an, for all γ N) ɛ)β 2 βa ) N 2 + N 2) 2β3 E ) ) 4 G G 3 βa ) N ) Proof. We procee as in the proof of Proposition 3.2 except that from the efinition of γ N) k, the prouct of transition probabilities insie the sum over j k in 3.8), is replace with 4.6). We use Lemma 4.3 instea of Lemma 3. to boun this sum. When N =, then also k = an γ ) is j p β ω 0) ) j = ω 0) β p 2 β ω 0) ω 0) ω ) j + I ) {ω l+ ω) =±e l } 2 j ) j = ω ) I ) {ω j j =ω 0) j i = 0 i l I ) {ω l+ ω) =±e l } 0 } 2 ) p ω0) ω) i ω ) i, ω ) i+ j i = 0 i l ) p ω0) ω) i ω ) i, ω ) i+ 4.7) where we have use the usual boun 2.5) an the fact that I ω ) ) j + {ω j + =ω) j ±e } observe that = 2. Now I {ω ) l+ ω) l =±e } 2 j i = 0 i l ) p ω0) ω) i ω ) i, ω ) i+ = Q l, ω 0) ωj = ω ) j ), 4.8) 0

11 so that 4.7) is equal to β 2 ω 0) p β ω 0) j ) I ) {ω j = ω ) j =ω 0) j = β p 2 β ω 0) ω 0) j ) j = 0) l, ω 0 }Q ωj = ω ) Q l, ω 0) ωj = ω 0) 0 ). Now use 4.3) with i = 0 to get the require boun. For the remaining cases we begin by ajusting the sum over j k in 3.8) as in 4.7) an 4.8). When N > an k = we use the same bouns as in the proof of Proposition 3.2 except that we use 4.3) with i = on the sum over j. This gives us a boun on γ N) when N > ) of 2β 2 ) 3 G 3 β G N i=2 j ) βa. 4.9) When N > an k = N, we use the same bouns as in the proof of Proposition 3.2 except that we use 4.3) with i = 0 on the sum over in 3.8). This gives us a boun on γ N) N when N > ) of β ) 2 G 2 β E ) N i=2 βa. 4.20) Similarly when N > an k N so N > 2) we use 4.3) on the sum over j k get a boun on γ N) k of the form in 3.8) to β G β E ) 2β 2 ) 3 N G 3 i=2 i k βa. 4.2) Simplifying these expressions an summing over k completes the proof of the lemma. Corollary 4.5 Summary of bouns). For all β [0, ], an such that a < N= N= N= ρ N) E 0) + G E ) ) a ), 4.22) χ N) E 0 ) + G E )2 a ) ) a ) 2, 4.23) γ N) G 2 ) 2 + ɛ) a + 2E )G G 3 ) 4 a ) ) Proof. Firstly note that the conition on a ensures that ρ N), χ N) an γ N) are all summable over N, an in all cases the supremum over β occurs at β = see Lemmas 4., 4.2 an 4.4). The results are then easily obtaine by summing each of the bouns in Lemmas 4., 4.2 an 4.4 over N.

12 5 Proof of Theorem. For such that a <, the bouns of Corollary 4.5 hol. From 4.7) we have the require absolute summability conitions in the iscussion after 2.2), an in particular 2.2) hols for all β [0, ]. To complete the proof of the theorem, it remains to show that the right han sie of 2.3) is no more than. By 4.7) an Corollary 4.5, we have boune times the right han sie of 2.3) by the sum of the right han sies of the bouns in Corollary 4.5. Since these terms all involve simple ranom walk Green s functions quantities, we will nee to use estimates of these quantities. In orer to boun E i ), we shall first prove that, for all i 0, E i ) = ) i+g i+). 5.) In orer to prove 5.), we first make use of [9, Lemma B.3], which states that G n x) is nonincreasing in x i for every i =,...,, so that the supremum in 3.3) can be restricte to v = 0 an v = e for any neighbour e of the origin. In orer to boun G n e), we make use of the fact that for any function x fx) for which fe) is constant for all e Z with e =, we have fe) = D f)0), so that { ) i+g i+) E i ) = max 0), ) } i+d G i+) )0). 5.2) Note that since G x) = δ 0,x + D G )x) δ 0,x, we have that G i 0) an G i+) 0) = G i 0) + D G i+) )0). Therefore, ) i+g i+) 0) = which is strictly larger than By [9, Lemma C.], G n ) i+d G i+) )0) + ) i+g i 0), 5.3) ) i+d G i+) )0) an thus proves 5.). is monotone ecreasing in for each n, so that it suffices to show that the sum of terms on the right han sies of 4.22), 4.23) an 4.24) is boune by for = 9. For this we use the following rigorous Green s functions estimates [8, 9] for = 8: G.07865, G 2.289, G ) Putting in these values for = 8 we get that the sum of the right han sies of the bouns in Corollary 4.5 is at most 0.97, whence the result follows for 9. To prove monotonicity for β [0, β 0 ] for some β 0 ) for each 8, it is sufficient to prove that χ ) < when 8 an that the other terms are boune), since this is the only term that oes not contain a factor β that can be mae arbitrarily small by choosing β 0 small. Since χ ) E 0 ), it is enough to show that E 0 ) < for = 8, since the right han sies of 4.22), 4.23) an 4.24) are boune for 8. From [9] we have 6G 5 5 < 6.57) <, an since 5 E 0 ) is ecreasing in, this completes the result. Acknowlegements. The work of RvH an MH was supporte in part by Netherlans Organisation for Scientific Research NWO). The work of MH was also supporte by a FRDF grant from the University of Aucklan. The authors thank Itai Benjamini for suggesting this problem to us, Takashi Hara for proviing the Green s functions upper bouns in 5.4), an an anonymous referee for many helpful suggestions that significantly improve the presentation of this paper. 2

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